# Derivative of $\sin^2 (\sqrt{t})$

I need to find the derivative of $\sin^2 (\sqrt{t})$ which I believe have done but the answer seems to be more simplified and I don't know how to arrive to it. Here are my steps

\begin{align} & \frac{d}{dt}(\sin \sqrt{t})^2 \\ & = 2(\sin \sqrt{t}) \cdot \frac{\cos \sqrt{t}}{2\sqrt{t}} \\ & = \frac{\sin \sqrt{t} \cdot \cos \sqrt{t}}{\sqrt{t}} \end{align}

The question is how does this simplify to $\frac{\sin 2\sqrt{t}}{2\sqrt{t}}$?

EDIT:

I forgot to add $2$ but I don't understand why top and bottom don't just cancel?

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$\sin 2\theta=2\sin\theta\cos\theta$. So this will simplify to what you say, divided by $2$. – 1015 Apr 9 '13 at 13:21
@Julien but I don't have $2\sin\theta\cos\theta$? The $2$ on the top and $2$ on the bottom cancels out. – Jeel Shah Apr 9 '13 at 13:22
$\frac{x}{y}=\frac{2x}{2y}$. Now you have the $2$ you miss at the numerator to apply the formula. Or if prefer, use $\sin\theta\cos\theta=\frac{\sin 2\theta}{2}$ directly. – 1015 Apr 9 '13 at 13:24
@RonGordon Sorry, I am a little confused. I plugged this into W|A and it has gives me the same simplified form that I arrived at. – Jeel Shah Apr 9 '13 at 13:26
In response to the edit: It'd be sin to cancel them out. The expression is $\dfrac{\sin(2\sqrt t)}{2\sqrt t}$ rather than something else. – Lord_Farin Apr 9 '13 at 13:34

Hint: If $y=x$ in the formula $\sin \left(x+y\right)=\sin x \cos y + \cos x \sin y$ then $$\sin(2x)= 2\cos(x)\sin(x).$$ Use this formula whit $x=\sqrt{t}$ to give \begin{align} \frac{\sin \sqrt{t} \cdot \cos \sqrt{t}}{\sqrt{t}}= & \frac{2\sin \sqrt{t} \cdot \cos \sqrt{t}}{2\sqrt{t}} \\ = & \frac{\sin (2\sqrt{t}) }{2\sqrt{t}} \end{align}
$$\frac{\sin \sqrt{t} \cdot \cos \sqrt{t}}{\sqrt{t}}=\frac{2\sin \sqrt{t} \cdot \cos \sqrt{t}}{2\sqrt{t}}=\frac{\sin (2\sqrt{t}) }{2\sqrt{t}}$$
Your answer is correct and the target answer is missing a factor of $2$ in the denominator. Here is confirmation from Alpha